The number of integral terms in the expansion | vedclass

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(D) The general term of the expansion is given by $T_{r+1} = {}^{1016}C_{r} (5^{\frac{1}{2}})^{1016-r} (7^{\frac{1}{8}})^{r}$.For the term to be an in

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$128$

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(D) The general term of the expansion is given by $T_{r+1} = {}^{1016}C_{r} (5^{\frac{1}{2}})^{1016-r} (7^{\frac{1}{8}})^{r}$.For the term to be an integer,the exponents of $5$ and $7$ must be integers.The exponent of $7$ is $\frac{r}{8}$,so $r$ must be a multiple of $8$.The exponent of $5$ is $\frac{1016-r}{2} = 508 - \frac{r}{2}$. For this to be an integer,$r$ must be even.Since $r$ must be a multiple of $8$,it is automatically even.Thus,$r \in \{0, 8, 16, \dots, 1016\}$.This is an arithmetic progression where $a = 0$,$d = 8$,and the last term $l = 1016$.Using the formula $l = a + (n-1)d$,we get $1016 = 0 + (n-1)8$.$n-1 = \frac{1016}{8} = 127$.$n = 128$.Therefore,there are $128$ integral terms.

The number of integral terms in the expansion of $(5^{\frac{1}{2}} + 7^{\frac{1}{8}})^{1016}$ is

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